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- Question 7, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- ntity ${{\cos }^{4}}\theta -{{\sin }^{4}}\theta =\dfrac{1}{\sec 2\theta }$. ====Solution==== \begin{align... quad \text{(By using double angle identity)}\\ &=\dfrac{1}{\sec 2\theta }=R.H.S.\end{align} =====Question 7(ii)===== Prove the identity $\tan \dfrac{\theta }{2}+co\operatorname{t}\dfrac{\theta }{2}=\dfrac{2}{\sin \theta }$. ====Solution==== \begin{align
- Question 2, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- ====Question 2(i)==== Evaluate exactly: $\sin \dfrac{\pi }{12}$ ===Solution=== We rewrite $\dfrac{\pi }{12}$ as $\dfrac{\pi }{3}-\dfrac{\pi }{4}$ and using the identity: \begin{align}\sin (\alpha -\beta )=\sin \alpha \cos \b
- Question 5, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- \circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}=\dfrac{1}{16}.$$ ====Solution==== We know that\\ $2\cos ... &=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2} \right)\cos {{80}^{\circ }}\\ &=\dfrac{1}{2}\cos {{80}^{\circ }}\cos {{40}^{\circ }}\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( 2\,\cos {{80}^{\circ }}\cos {{40}^{\c
- Question 5, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- \circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}=\dfrac{1}{16}$. ====Solution==== We know that\\ $2\cos \... &=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2} \right)\cos {{80}^{\circ }}\\ &=\dfrac{1}{2}\cos {{80}^{\circ }}\cos {{40}^{\circ }}\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( 2\,\cos {{80}^{\circ }}\cos {{40}^{\c
- Question 6, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- c }}$. ====Solution==== Because ${{15}^{\circ }}=\dfrac{{{30}^{\circ }}}{2}$, and $\dfrac{\theta }{2}=\dfrac{{{30}^{\circ }}}{2}$, we can find $\cos {{15}^{\circ }}$by using half angle identity as, \begin{align}\cos {{15}^{\circ }}&=\cos \dfrac{{{30}^{\circ }}}{2}=\sqrt{\dfrac{1+\cos {{30}^{\c
- Question 3, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- ar, Pakistan. =====Question 3(i)===== If $\sin u=\dfrac{3}{5}$ and $\sin v=\dfrac{4}{5}$ where$u$ and $v$ are between $0$ and $\dfrac{\pi }{2}$, evaluate each of the following exactly ... ft( u+v \right)$ ====Solution==== Given $\sin u=\dfrac{3}{5},$ $0\le u\le \dfrac{\pi }{2}.$ $\sin v
- Question 7, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- (i)===== Separate into real and imaginary parts $\dfrac{2+3i}{5-2i}$. ====Solution==== \begin{align}&\dfrac{2+3i}{5-2i} \\ =&\dfrac{2+3i}{5-2i}\times \dfrac{5+2i}{5+2i} \quad \text{by rationalizing} \\ =&\dfrac{10-6+15i+4i}{25+4}\\ =&\d
- Question 8, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- ===Question 8(i)===== Prove that: $\tan \left( \dfrac{\pi }{4}+\theta \right)=\dfrac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }$ ====Solution==== \begin{align}L.H.S.&=\tan \left( \dfrac{\pi }{4}+\theta \right)\\ &=\dfrac{\sin \left( \dfrac{\pi }{4}+\theta \right)}{\cos \left( \dfrac{\pi
- Question, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- stan. =====Question 4(i)===== If $\sin \alpha =-\dfrac{4}{5}$ and $\cos \beta =-\dfrac{12}{13}$, $\alpha $in Quadrant III and $\beta $in Quadrant II, find th... \right)$. ====Solution==== Given: $\sin \alpha=-\dfrac{4}{5}$, $\alpha$ is in 3rd quadrant, \\ $\sin \beta=-\dfrac{12}{13}$, $\beta$ is in 2nd quadrant. We have an
- Question11 and 12, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- the angles of a triangle $ABC$, show that $\cot \dfrac{\alpha }{2}+\cot \dfrac{\beta }{2}+\cot \dfrac{\gamma }{2}=\cot \dfrac{\alpha }{2}\cot \dfrac{\beta }{2}\cot \dfrac{\gamma }{2}$ ====Solution==== Sinc
- Question 2, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- dentity: $$\sin \alpha +\sin \beta =2\sin \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right).$$ Put $\alpha ={{37}^{\circ }}$... 37}^{\circ }}+\sin {{43}^{\circ }}&=2\sin \left( \dfrac{{{37}^{\circ }}+{{43}^{\circ }}}{2} \right)\cos \left( \dfrac{{{37}^{\circ }}-{{43}^{\circ }}}{2} \right)\\
- Question 6, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- ign}{{z}^{4}}+{{z}^{2}}+1&=0\\ {{z}^{4}}+2\left( \dfrac{1}{2} \right){{z}^{2}}+\dfrac{1}{4}-\dfrac{1}{4}+1&=0\\ {{\left( {{z}^{2}}+\dfrac{1}{2} \right)}^{2}}+\dfrac{4-1}{4}&=0\\ {{\left( {{z}^{2}}+\dfrac
- Question 8 & 9, Review Exercise 10 @fsc-part1-kpk:sol:unit10
- =Question 8===== Prove the identity $\sin \left( \dfrac{\pi }{4}-\theta \right)\sin \left( \dfrac{\pi }{4}+\theta \right)=\dfrac{1}{2}\cos 2\theta $. ====Solution==== We know that $2\sin \alpha \sin \b... beta \right)$ \begin{align}L.H.S.&=\sin \left( \dfrac{\pi }{4}-\theta \right)\sin \left( \dfrac{\pi }{
- Question 4 & 5, Review Exercise 10 @fsc-part1-kpk:sol:unit10
- Question 4===== Prove the identity ${{\sin }^{2}}\dfrac{\theta }{2}=\dfrac{\sin \theta \tan \dfrac{\theta }{2}}{2}$. ====Solution==== \begin{align}R.H.S.&=\dfrac{\sin \theta \tan \dfrac{\theta }{2}}{2}\\ &=\dfra
- Question 8, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- , Pakistan. =====Question 8(i)===== Express the $\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}$ in the standard form of $a+ib.$ ====Solution==== \begin{align}&\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}\\ &=\dfrac{\left( 3+2i \right)\left( 1-2i \right)+\left( 2+i \right)\left(